(a) Explanation The blocks are placed on the frictionless surface therefore; the system will accelerate in the direction of the heavier mass with acceleration a . The free body diagram for 3.50 kg mass is given as. Write the expression for equilibrium of the block as. T − m 1 g sin θ = m 1 a (I) Here, T is the tension developed in the string, m 1 is the mass of the lighter block, g is the acceleration due to gravity, a is the acceleration of the block along the incline and θ is the angle made by the incline. The free body diagram for 3.50 kg mass is given as. Write the expression for equilibrium of the block as. m 2 g sin θ − T = m 2 a (II) Conclusion: Substitute 3.50 kg for m 1 , 9.80 m / s 2 for g and 35 ° for θ in equation (I). T − ( 3.50 kg ) ( 9.80 m / s 2 ) sin 35 ° = ( 3 (b) To determine The tension in the string.

Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses)

9th Edition

ISBN: 9781305266292

Author: Raymond A. Serway, John W. Jewett

Publisher: Cengage Learning

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Chapter 5, Problem 49P

(a)

To determine

The magnitude of acceleration of each block.

(b)

To determine

The tension in the string.

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Chapter 5, Problem 48P

Chapter 5, Problem 50P

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Chapter 5 Solutions

Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses)

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The magnitude of acceleration of each block.

Solution Summary: The author explains the magnitude of acceleration of each block and the free body diagram for 3.50kg mass.

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The tension in the string.

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Chapter 5 Solutions